Let $g$ be a function defined for all real numbers except for $-3$. Also let $g'$, the derivative of $g$, be defined as $g'(x)=\dfrac{x}{x+3}$. On which intervals is $g$ decreasing? Choose 1 answer: Choose 1 answer: (Choice A) A $x>0$ only (Choice B) B $x<-3$ and $-3<x<0$ (Choice C) C $-3<x<0$ only (Choice D) D $x<-3$ only (Choice E) E The entire domain of $g$
Answer: We can analyze the intervals where $g$ is increasing/decreasing by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $g'(x)=\dfrac{x}{x+3}$ and that $g$ is undefined at $x=-3$. $g'(x)=0$ for $x=0$. Our critical point is $x=0$, and we should also consider $x=-3$. Our points divide the number line into three intervals: $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $\begin{array}{rl} x<& \llap{-}3\end{array}$ $ \llap{-}3<x<0$ $ x>0$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<-3$ $x=-4$ $g'(-4)=4>0$ $g$ is increasing $\nearrow$ $-3<x<0$ $x=-2$ $g'(-2)=-2<0$ $g$ is decreasing $\searrow$ $x>0$ $x=1$ $g'(1)=\dfrac{1}{4}>0$ $g$ is increasing $\nearrow$ In conclusion, $g$ is decreasing over the interval $-3<x<0$ only.